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          Leetcode2035 将数组分成两个数组并最小化数组和的差
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              <time title="创建时间：2021-10-18 16:32:23 / 修改时间：20:59:25" itemprop="dateCreated datePublished" datetime="2021-10-18T16:32:23+08:00">2021-10-18</time>
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        <h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/partition-array-into-two-arrays-to-minimize-sum-difference/">原题链接</a><br>给定一个长度为2*n的数组，将这个数组分成两个长度为n的数组，分别求出两个数组之和，并最小化他们的差值，返回差值的绝对值。</p>
<span id="more"></span>
<p><img src="image-20211018203146120.png" alt="image-20211018203146120"></p>
<p><img src="image-20211018203204379.png" alt="image-20211018203204379"></p>
<p>数据范围：</p>
<p><img src="image-20211018203319493.png" alt="image-20211018203319493"></p>
<h2 id="分析："><a href="#分析：" class="headerlink" title="分析："></a>分析：</h2><p>n&lt;=15，因此可以接收复杂度为指数。但是数组的长度为2n，因此不能对全部数组进行枚举。<br>这里使用到一个方法：分治，具体来讲就是<strong>折半查找</strong>。</p>
<ul>
<li>将数组分为左右两个部分</li>
<li>分别枚举左右两部分的子集的和，共2的n次方种可能</li>
<li>上一步得到的其实是一个二维数组，f(i,j)，其中i表示有i个数字，f(i,)表示从左边(右边)选取i个数字，所有可能的和。</li>
<li>为了得到答案，我们先对right的二维数组中每一行进行排序</li>
<li>左边选k个，则右边选n-k个，则同时选了n个，剩下n个，求abs即可</li>
<li>为了得到最优解，选取的数和留下的数尽量靠近sum/2</li>
<li>遍历left，二分查找right<h2 id="代码实现"><a href="#代码实现" class="headerlink" title="代码实现"></a>代码实现</h2>为了枚举所有子集，使用状态压缩。<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">minimumDifference</span><span class="params">(<span class="keyword">int</span>[] nums)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// HashMap&lt;Integer,List&lt;Integer&gt; &gt;left=new HashMap&lt;&gt;();</span></span><br><span class="line">        <span class="comment">// HashMap&lt;Integer,List&lt;Integer&gt; &gt;right=new HashMap&lt;&gt;();</span></span><br><span class="line">        List&lt;List&lt;Integer&gt; &gt;left=<span class="keyword">new</span> ArrayList&lt;&gt;();</span><br><span class="line">        List&lt;List&lt;Integer&gt; &gt;right=<span class="keyword">new</span> ArrayList&lt;&gt;();</span><br><span class="line"></span><br><span class="line">        <span class="keyword">int</span> len=nums.length;</span><br><span class="line">        <span class="keyword">int</span> n=len/<span class="number">2</span>;</span><br><span class="line">        <span class="keyword">int</span> []a=<span class="keyword">new</span> <span class="keyword">int</span>[n];</span><br><span class="line">        <span class="keyword">int</span> []b=<span class="keyword">new</span> <span class="keyword">int</span>[n];</span><br><span class="line">        <span class="keyword">int</span> sum=<span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;n;i++)&#123;</span><br><span class="line">            a[i]=nums[i];</span><br><span class="line">            b[i]=nums[i+n];</span><br><span class="line">            sum=sum+a[i]+b[i];</span><br><span class="line">            left.add(<span class="keyword">new</span> ArrayList&lt;&gt;());</span><br><span class="line">            right.add(<span class="keyword">new</span> ArrayList&lt;&gt;());</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">//单独加一次是因为可以选0个数字到n个数字，一共n+1个选项</span></span><br><span class="line">        left.add(<span class="keyword">new</span> ArrayList&lt;&gt;());</span><br><span class="line">        right.add(<span class="keyword">new</span> ArrayList&lt;&gt;());</span><br><span class="line">        <span class="comment">// 分割数组，计算sum</span></span><br><span class="line">        <span class="comment">// 枚举子集，每一个都有2^n个状态</span></span><br><span class="line">        <span class="comment">// 状态压缩</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> mask=<span class="number">0</span>;mask&lt;<span class="number">1</span>&lt;&lt;n;mask++)&#123;</span><br><span class="line">            <span class="keyword">int</span> key = Integer.bitCount(mask);<span class="comment">//统计有多少个二进制1</span></span><br><span class="line">            <span class="keyword">int</span> cnta=<span class="number">0</span>;</span><br><span class="line">            <span class="keyword">int</span> cntb=<span class="number">0</span>;</span><br><span class="line">            <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;n;i++)&#123;</span><br><span class="line">                <span class="keyword">if</span> (((mask &gt;&gt; i) &amp; <span class="number">1</span>) == <span class="number">1</span>) &#123;</span><br><span class="line">                    cnta += a[i];</span><br><span class="line">                    cntb += b[i];</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">            left.get(key).add(cnta);</span><br><span class="line">            right.get(key).add(cntb);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">//将right进行sort，方便二分查找</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;=n;i++)&#123;</span><br><span class="line">            Collections.sort(right.get(i));</span><br><span class="line">        &#125; </span><br><span class="line">        <span class="comment">// 开始二分</span></span><br><span class="line">        <span class="keyword">int</span> res=Integer.MAX_VALUE;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> k=<span class="number">0</span>;k&lt;=n;k++)&#123;</span><br><span class="line">            List&lt;Integer&gt;templeft=left.get(k);</span><br><span class="line">            List&lt;Integer&gt;tempright=right.get(n-k);</span><br><span class="line">            <span class="keyword">for</span>(<span class="keyword">int</span> t:templeft)&#123;</span><br><span class="line">                <span class="keyword">int</span> lo=<span class="number">0</span>;</span><br><span class="line">                <span class="keyword">int</span> hi=tempright.size();</span><br><span class="line">                <span class="keyword">while</span>(lo&lt;hi)&#123;</span><br><span class="line">                    <span class="keyword">int</span> mid=(lo+hi)/<span class="number">2</span>;</span><br><span class="line">                    <span class="keyword">int</span> j=tempright.get(mid);</span><br><span class="line">                    <span class="keyword">if</span>(j+t&lt;sum/<span class="number">2</span>)lo=mid+<span class="number">1</span>;</span><br><span class="line">                    <span class="keyword">else</span> hi=mid;</span><br><span class="line">                &#125;</span><br><span class="line">                <span class="comment">// System.out.println(k+&quot; &quot;+templeft+&quot; &quot;+tempright);</span></span><br><span class="line">                  <span class="keyword">if</span>(lo&gt;=tempright.size())lo--;</span><br><span class="line">                res = Math.min(res, Math.abs((sum - t - tempright.get(lo) - (t + tempright.get(lo)))));</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
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